How do you evaluate ${sin}^{- 1} (sin (\frac{19 π}{10}))$ $sin^-1(sin((19pi)/10))$ ?

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How do you evaluate ${sin}^{- 1} (sin (\frac{19 π}{10}))$ $sin^-1(sin((19pi)/10))$ ?

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Answer 1 · 2017-07-26T20:15:50

$- \frac{π}{10}$ $-pi/10$

Explanation:

${sin}^{- 1} (x)$ $sin^-1(x)$ is a number (or angle), $t$ $t$ in $[- \frac{π}{2}, \frac{π}{2}]$ $[-pi/2,pi/2]$
with $sin (t) = x$ $sin(t) = x$

So,

${sin}^{- 1} (sin (\frac{19 π}{10}))$ $sin^-1(sin((19pi)/10))$ is a number (or angle), $t$ $t$ in $[- \frac{π}{2}, \frac{π}{2}]$ $[-pi/2,pi/2]$
with $sin (t) = sin (\frac{19 π}{10})$ $sin(t) = sin((19pi)/10)$

$\frac{19 π}{10}$ $(19pi)/10$ is almost $2 π$ $2pi$ , so it is a fourth quadrant angele.

The reference angle for $\frac{19 π}{10}$ $(19pi)/10$ is $\frac{π}{10}$ $pi/10$ .

So $t = - \frac{π}{10}$ $t = -pi/10$

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How do you evaluate ${sin}^{- 1} (sin (\frac{19 π}{10}))$ $sin^-1(sin((19pi)/10))$ ?

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Explanation:

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How do you evaluate ${sin}^{- 1} (sin (\frac{19 π}{10}))$ $sin^-1(sin((19pi)/10))$ ?